Question

If `13/4 L` of a gas at room temperature exerts a pressure of `16 kPa` on its container, what pressure will the gas exert if the container's volume changes to `5/6 L`?

Answer 1

The pressure is `=62.4 kPa`

Explanation:

We apply Boyle's law.

`P_1V_1=P_2V_2`

`P_1=16kPa`

`V_1=13/4l`

`V_2=5/6l`

So `P_2=(P_1V_1)/V_2=(16*13/4)/(5/6)=4*13*6/5=312/5=62.4 kPa`